4 strike through incorrect assertion.

Different take: For the application you want, however, you don't need such a general theory. In your application, the order of every element divides the order of the distinguished element. The disadvantage is that this cannot be expressed as an identity, so you do not have a variety of algebras. You do not have the nice constructions such as coproducts, products, free objects, etc. a priori. On the other hand, I think that the classification is pretty straightforward: every such group can be expressed as a product of cyclic groups, $C_{n_1}\times\cdots\times C_{n_k}$, with $n_1|n_2|\cdots|n_k$, and with distinguished element a specific generator of the largest cyclic factor in the case where $n_k>0$, and an arbitrary nonzero element of the last cyclic factor if $n_k=0$ and a torsion element plus a nonzero element of the last cyclic factor if $n_k=0$. However, this decomposition is not decomposing the objects into something which is a direct product of pointed abelian groups (the direct summands other than the largest ones do not contain the distinguished element). The proof in the finite case can be done the usual way; for an abelian $p$-group $A$, if $x$ is an element of maximal order then $A$ is isomorphic to $A/\langle x\rangle \oplus \langle x\rangle$, so you can start by picking the distinguished element; otherwise, consider the $p$-parts and think of them as quotients of the original pointed group, with distinguished element the image of the distinguished element; then put them together in the usual way, and the distinguished element will correspond to a generator of the largest cyclic factor. In the infinite case, it gets a bit more complicated, but it should go through: take $A/A^{tor}$, with distinguished element the image of the distinguished element of $A$; this is a direct sum of cyclic groups of infinite order as an abelian group, and the subgroup generated by the distinguished element is a submodule of rank $1$. So if I remember my modules over PIDs correctly, there is a basis for $A/A^{tor}$ of the form $x_1,\ldots,x_r$, in which $1$ is a multiple of $x_r$; then deal with the torsion part as you would with a normal finite abelian group. Edit: No, the infinite case does not go through like this; see comment. You can get it so that it will be of the form $(a,(0,\ldots,0,d))$, where $a$ is in the torsion part and $(0,\ldots,0,d)$ is an element of the torsion free part with $d>0$, but you may not be able to "get rid" of the $a$. This is problematical.

3 fix silly mistake on decomposing an abelian group

Different take: For the application you want, however, you don't need such a general theory. In your application, the order of every element divides the order of the distinguished element. The disadvantage is that this cannot be expressed as an identity, so you do not have a variety of algebras. You do not have the nice constructions such as coproducts, products, free objects, etc. a priori. On the other hand, I think that the classification is pretty straightforward: every such group can be expressed as a product of cyclic groups, $C_{n_1}\times\cdots\times C_{n_k}$, with $n_1|n_2|\cdots|n_k$, and with distinguished element a specific generator of the largest cyclic factor . in the case where $n_k>0$, and an arbitrary nonzero element of the last cyclic factor if $n_k=0$. However, this decomposition is not decomposing the objects into something which is a direct product of pointed abelian groups (the direct summands other than the largest ones do not contain the distinguished element). The proof in the finite case can be done the usual way, since if you have a finitely generated ; for an abelian group, and $p$-group $A$, if $x$ is an element of maximal order of $A$, then $A$ is isomorphic to $A/\langle x\rangle \oplus \langle x\rangle$, so you can start by picking the distinguished element; otherwise, consider the $p$-parts and think of them as quotients of the original pointed group, with distinguished element the image of the distinguished element; then put them together in the usual way, and the distinguished element will correspond to a generator of the largest cyclic factor. So In the infinite case, it gets a bit more complicated, but it should go through: take $A/A^{tor}$, with distinguished element the image of the distinguished element of $A$; this is a direct sum of cyclic groups of infinite order as an abelian group, and the subgroup generated by the distinguished element first and go from is a submodule of rank $1$. So if I remember my modules over PIDs correctly, there is a basis for $A/A^{tor}$ of the form $x_1,\ldots,x_r$, in which $1$ is a multiple of $x_r$; then deal with the torsion part as you would with a normal finite abelian group.

2 fix the 1-generator pointed case.

This category includes the entire theory of abelian groups as the class of those groups in which $0=1$; so any classification theory will be much harder than the one for abelian groups. For example, the $1$ generator pointed abelian groups: every $1$-generator pointed abelian group is really a $2$-generator abelian group: it is generated by the given generator plus the distinguished element $1$. So suppose $x$ is the given generator. If $\langle 1\rangle\subseteq\langle x\rangle$, then you get one distinct an isomorphism class for each pair type of the form $(C,C')$, where $C$ is a cyclic group and $C'$ is a subgroup of $C$ (corresponding $C$ corresponds to the subgroup cyclic group generated by $x$, and $C'$ to the distinguished element; abelian subgroup generated by $1$). But these do not exhaust all possibilities. You also get groups correspond in which $\langle x\rangle\cap\langle 1\rangle = \{0\}$; if the orders are coprime, you are back to the previous case$C'=0$). , but if the orders are not coprime then the description is a bit more complicated. And then the case where they intersect nontrivially but neither contains the other, and so on. As you can see, things get complicated even in this simplest of cases...

1